009B Sample Midterm 1, Problem 4

Evaluate the integral:

$\int \sin ^{3}x\cos ^{2}x~dx$

Foundations:
1. Recall the trig identity
$\sin ^{2}x+\cos ^{2}x=1$
2. How would you integrate  $\int \sin ^{2}x\cos x~dx?$
You can use  $u$-substitution.
Let  $u=\sin x.$
Then,  $du=\cos x~dx.$
Thus,
${\begin{array}{rcl}\displaystyle {\int \sin ^{2}x\cos x~dx}&=&\displaystyle {\int u^{2}~du}\\&&\\&=&\displaystyle {{\frac {u^{3}}{3}}+C}\\&&\\&=&\displaystyle {{\frac {\sin ^{3}x}{3}}+C.}\end{array}}$

Solution:

Step 1:
First, we write
$\int \sin ^{3}x\cos ^{2}x~dx=\int (\sin x)\sin ^{2}x\cos ^{2}x~dx.$
Using the identity  $\sin ^{2}x+\cos ^{2}x=1,$
we get
$\sin ^{2}x=1-\cos ^{2}x.$
If we use this identity, we have
${\begin{array}{rcl}\displaystyle {\int \sin ^{3}x\cos ^{2}x~dx}&=&\displaystyle {\int (\sin x)(1-\cos ^{2}x)\cos ^{2}x~dx}\\&&\\&=&\displaystyle {\int (\cos ^{2}x-\cos ^{4}x)\sin(x)~dx.}\end{array}}$

Step 2:
Now, we use  $u$-substitution.
Let  $u=\cos(x).$
Then,  $du=-\sin(x)dx.$
Therefore,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\int\sin^3x\cos^2x~dx} & = & \displaystyle{\int -(u^2-u^4)~du}\\ &&\\ & = & \displaystyle{\frac{-u^3}{3}+\frac{u^5}{5}+C}\\ &&\\ & = & \displaystyle{\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C.} \end{array}}

Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C}

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